作业帮已知函数f(x)=log2^ ( x/4 ) ×log2^ (2x) (1)解不等式f(x)>0;(2)当x∈【1,4】时,求f(x)的值域f（x）=log2（2x）×log2（x／4）＝[（log2 2）+(log2 x)] ×[(log2 x) -(log2 4)]=[1+(log2 x)] ×[(log2 x) -2]=(log2 x)² - (log2 x) -2
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![已知函数f(x)=log2^ ( x/4 ) ×log2^ (2x) (1)解不等式f(x)>0;(2)当x∈【1,4】时,求f(x)的值域f（x）=log2（2x）×log2（x／4）＝[（log2 2）+(log2 x)] ×[(log2 x) -(log2 4)]=[1+(log2 x)] ×[(log2 x) -2]=(log2 x)² - (log2 x) -2](/uploads/image/z/4344891-51-1.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dlog2%5E+%28+x%2F4+%29+%C3%97log2%5E+%282x%29+%281%29%E8%A7%A3%E4%B8%8D%E7%AD%89%E5%BC%8Ff%28x%29%3E0%3B%282%29%E5%BD%93x%E2%88%88%E3%80%901%2C4%E3%80%91%E6%97%B6%2C%E6%B1%82f%28x%29%E7%9A%84%E5%80%BC%E5%9F%9Ff%EF%BC%88x%EF%BC%89%3Dlog2%EF%BC%882x%EF%BC%89%C3%97log2%EF%BC%88x%EF%BC%8F4%EF%BC%89%EF%BC%9D%5B%EF%BC%88log2+2%EF%BC%89%2B%28log2+x%29%5D+%C3%97%5B%28log2+x%29+-%28log2+4%29%5D%3D%5B1%2B%28log2+x%29%5D+%C3%97%5B%28log2+x%29+-2%5D%3D%28log2+x%29%26%23178%3B+-+%28log2+x%29+-2)
已知函数f(x)=log2^ ( x/4 ) ×log2^ (2x) (1)解不等式f(x)>0;(2)当x∈【1,4】时,求f(x)的值域f（x）=log2（2x）×log2（x／4）＝[（log2 2）+(log2 x)] ×[(log2 x) -(log2 4)]=[1+(log2 x)] ×[(log2 x) -2]=(log2 x)² - (log2 x) -2
已知函数f(x)=log2^ ( x/4 ) ×log2^ (2x) (1)解不等式f(x)>0;(2)当x∈【1,4】时,求f(x)的值域
f（x）=log2（2x）×log2（x／4）＝[（log2 2）+(log2 x)] ×[(log2 x) -(log2 4)]=[1+(log2 x)] ×[(log2 x) -2]=(log2 x)² - (log2 x) -2=[(log2 x) -1/2]²-9/4因为x∈[1／2,4]所以(log2 x) ∈[-1,2]则当(log2 x)=1/2即x＝－1时,函数有最小值为-9/4当(log2 x)=－1或2即x＝1/2或4时,函数有最大值为0 .我的问题是.X属于【1/2,4】是怎么变成（log2x)属于【-1,2】?
已知函数f(x)=log2^ ( x/4 ) ×log2^ (2x) (1)解不等式f(x)>0;(2)当x∈【1/2.4]时,求f(x)的值域
已知函数f(x)=log2^ ( x/4 ) ×log2^ (2x) (1)解不等式f(x)>0;(2)当x∈【1,4】时,求f(x)的值域f（x）=log2（2x）×log2（x／4）＝[（log2 2）+(log2 x)] ×[(log2 x) -(log2 4)]=[1+(log2 x)] ×[(log2 x) -2]=(log2 x)² - (log2 x) -2
由于log2 x是递增函数,故x越大,log2 x越大,故把1/2带入为-1,把4带入为2.故属于[-1,2].